Logic equivalence question

Question

I have a question about discrete mathematics question that I have been struggling to solve. Here is the question:

Each of the two rooms (room I and room II) contains either a lady or a tiger. If a room contains a lady, the sign on its door is true. If it contains a tiger, the sign is false. The signs are Room 1 - It makes no difference which room to pick. Room 2 - There is a lady in the other room. which room contains ladies? Use logic equivalences or rules of inference.

The only steps I could achieve were: I let p - there is a lady in room 1 q - there is a lady in room 2 and I said that sign 1 represents: (p ^ q) ∨ (~p ^ ~q) and I also said sign 2 represents: q ↔ p I don't know if it's correct but please provide your guidance. Thank you🙏

Answer 1

You got a good start.

Yes, sign 2 is saying $p$, and since the sign is true if and only if there is a lady in it, the information that you have with regard to room 2 is indeed: $$q \leftrightarrow p$$

For room 1 you made a small mistake. Yes, you got the right idea for symbolizing what the sign for room 1 is saying, which is that either both rooms contain ladies or both rooms contain tigers, which you symbolized correctly as $(p \land q) \lor (\neg p \land \neg q)$. However, again you need to use the information that what the sign is saying is true if and only if there is a lady in the room, meaning that the information you have with regard to room 1 is: $$p \leftrightarrow ((p \land q) \lor (\neg p \land \neg q))$$

OK, so now you need to combine these two pieces of information to get the result. (Hint: think of what makes $q \leftrightarrow p$ true). Good luck!

Answer 2

the trick is to take 2 cases that lady is in room $1$ or $2$, and then follow the arguments given in the question for both cases, and arrive at a contradiction in any one of the cases...

using propositional logic ( its enough )

let $r_1$ be room $1$ and $r_2$ be room $2$.

Let $\mathbf L \ $ be lady $\ \ $ and $ \ \ \ S_1 , S_2$ be signs on room $1$ and room $2$ respectively

let $\mathbf P = $ L is in $r_1$ ,$ \ \ \ \ \ \ \ \ \\ \ \ \ \ \ \ \ $ let $\mathbf Q = $ L is in $r_2$

let $\mathbf A = $ $S_1$ is true,$ \ \ \ \ \ \ \ \ \\ \ \ \ \ \ \ \ $ let $\mathbf B = $ $S_2$ is true

its given that $ ( P \rightarrow \lnot Q ) \ \land \ ( Q \rightarrow \lnot P )$ .... $(a)$

now, $ \ \ \ \bf CASE \ \ 1$

$(1)$ $\quad$ if $ \quad$ Q is true ,

$(2)$ then $\ \ \ \ $ Q $\rightarrow $ B, $\ \ \ \therefore S_2$ is true

$(3)$ but its given that $B \rightarrow P $, $\ \ \ \therefore P$ is true

$\therefore \ \text{from} \ \ (a), (1) \ \& \ (3)$ .. $\bf Q$ leads to a contradiction... $AND \ \ \ \bf P$ must be $\bf TRUE$.

( btw, you already knew lady was in room$1$, I believe )